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$$ \int_{-\infty}^{\infty}x\exp(-\vert x\vert) \sin(ax)\,dx\quad\mbox{where}\ a\ \mbox{is a positive constant.} $$ My idea is to use integration by parts. But I have been not handle three terms.. Please, help me solve that.

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    $\begingroup$ How about differentiating $$2\int_{0}^{\infty} \exp(-x)\cos(ax) \, dx$$w.r.t. $a$? $\endgroup$ – Sangchul Lee Sep 17 '16 at 0:42
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or note $$ 2\int_{0}^{\infty}x\exp(-x) \sin(ax)\,dx = 2\;\mathrm{Im}\int_{0}^{\infty}x\exp((-1+ia)x)\;dx $$

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  • $\begingroup$ Just like mine, so of course I will upvote it. $\endgroup$ – marty cohen Sep 18 '16 at 1:47
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Since the function is symmetric, $I(a) =\int_{-\infty}^{\infty}x\exp(-\vert x\vert) \sin(ax)\,dx =2\int_{0}^{\infty}x\exp(-x) \sin(ax)\,dx $.

At this point, I would write $\sin(ax) =Im(e^{iax}) $ so that $I(a) =2Im\int_{0}^{\infty}x\exp(-x) e^{iax}\,dx =2Im\int_{0}^{\infty}x\exp(x(-1+ia))dx $ and try to evaluate $\int_{0}^{\infty}x\exp(-bx)dx $, which should not be too difficult.

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  • $\begingroup$ All too easy. +1 .... $\endgroup$ – Mark Viola Sep 17 '16 at 1:54

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